What are some ways to prove $$\int_0^1 x^n dx=\frac{1}{n+1}$$ straight from (any) definition of the Riemann-/regulated integral? Or do we need the fundamental theorem of calculus and (anti)derivatives for that?

  • $\begingroup$ Why would you want to use the Riemann (or any other) definition? Normally you would evaluate this integral by the fundamental theorem of calculus. $\endgroup$ – uniquesolution Oct 10 '15 at 23:36
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    $\begingroup$ Try to prove by induction if $1^k+2^k...n^k=f(n)$ then f(n) is a polynomial with degree $k+1$ and leading coefficient $\frac{1}{k+1}$ $\endgroup$ – user175968 Oct 10 '15 at 23:38
  • $\begingroup$ Normally I would. Still I find the question interesting. $\endgroup$ – Damian Reding Oct 10 '15 at 23:38
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    $\begingroup$ If you know Faulhaber's formula, it' immediate from the definition of the Riemann integral. $\endgroup$ – Bernard Oct 10 '15 at 23:40
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    $\begingroup$ You can also apply Stoltz theorem to the Riemann sum of your integral. $\endgroup$ – Maciej Oct 10 '15 at 23:56

Use Someone's Trick: Consider a partition of $[0,1]$ defined by a geometric series.

Say $0<r<1$. We're going to consider an "infinite Riemann sum" corresponding to the "infinite partition" $$1>r>r^2\dots.$$

(Of course that's not quite kosher. We leave it to you to adjust what's below to work with an actual partition $1>r>\dots>r^N>0$, where $N$ is large enough that $r^N$ is very small. The formulas look so much nicer for the infinite partition...)

A super-Riemann sum corresponding to that partition is $$\sum_{j=0}^\infty f(r^j)(r^j-r^{j+1})$$(where $f(t)=t^n$). Factor out the $1-r$ and you're left with a geometric series; add the series and you see the sum is equal to$$\frac{1-r}{1-r^{n+1}}=\frac1{1+r+\dots+r^n}\to\frac1{n+1}\quad(r\to1).$$

  • $\begingroup$ Another trick that's really good to know :-) $\endgroup$ – Damian Reding Oct 11 '15 at 14:47

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